Chapter 13. Chemical Equilibrium

Transcription

1 Chapter 13 Chemical Equilibrium

2 Chapter 13 Preview Chemical Equilibrium The Equilibrium condition and constant Chemical equilibrium, reactions, constant expression Equilibrium involving Pressure Chemical expressions involving gases Heterogeneous Equilibria Application of equilibrium constant and its calculations for pressure and concentrations Le Châtellier s Principle Effect of a change of concentration, pressure, and temperature on the reaction equilibrium

3 Introduction Reactions that stop far short from completion and have concentrations of reactants and products remain constant with time has reached Chemical equilibrium. Equilibrium is a state in which there are no observable changes as time goes by. Chemical equilibrium is achieved when: reaction is carried out in a closed vessel the concentrations of the reactants and products remain constant Types of Equilibria: Physical equilibrium H O (l) Chemical equilibrium N O 4 (g) H O (g) NO (g)

4 13.1 Equilibrium conditions and constant N O 4 (g) NO (g) equilibrium equilibrium equilibrium Start with NO Start with N O 4 Start with NO & N O 4

5 13. The Equilibrium constant constant N O 4 (g) K = [NO ] [N O 4 ] aa + bb NO (g) = 4.63 x 10-3 cc + dd K = [C]c [D] d [A] a [B] b Law of Mass Action If K >> 1 Equilibrium Will Lie to the right Favor products K << 1 Lie to the left Favor reactants

6 Homogenous equilibrium applies to reactions in which all reacting species are in the same phase. CH 3 COOH (aq) + H O (l) CH 3 COO - (aq) + H 3 O + (aq) K c = [CH 3 COO - ][H 3 O + ] [CH 3 COOH][H O] [H O] = constant K c = [CH 3 COO- ][H 3 O + ] [CH 3 COOH] = K c [H O] General practice not to include units for the equilibrium constant.

7 Review Question 1 Consider the chemical system CO + Cl COCl ; K = L/mol. How do the equilibrium concentrations of the reactants compare to the equilibrium concentration of the product? 1) They are much smaller. ) They are much bigger. 3) They are about the same. 4) They have to be exactly equal. 5) You can t tell from the information given.

8 Review Question Determine the equilibrium constant for the system N O 4 NO at 5 C. The concentrations are shown here: [N O 4 ] = 9.43 X 10 M, [NO ] = 1.41 X 10 M a) b) 6.69 c) 474 d) 0.04 e)

9 13.3 The Equilibrium expression Involving Pressures K c = [NO ] [N O 4 ] N O 4 (g) NO (g) PV = nrt P = (n/v)rt P = CRT K p = P NO P N O 4 In most cases K c K p aa (g) + bb (g) cc (g) + dd (g) K p = K c (RT) n n = moles of gaseous products moles of gaseous reactants = (c + d) (a + b)

10 The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form COCl (g) at 74 0 C are [CO] = 0.01 M, [Cl ] = M, and [COCl ] = 0.14 M. Calculate the equilibrium constants K c and K p. CO (g) + Cl (g) COCl (g) K c = [COCl ] 0.14 = [CO][Cl ] 0.01 x = 0 K p = K c (RT) n n = 1 = -1 R = T = = 347 K K p = 0 x (0.081 x 347) -1 = 7.7 The equilibrium constant K p for the reaction NO (g) NO (g) + O (g) is 158 at 1000K. What is the equilibrium pressure of O if the P = atm and P NO NO = 0.70 atm? K p = P NO P O P NO P O P NO = K p P NO P O = 347 atm

11 13.4 Heterogeneous Equilibria Heterogenous equilibrium applies to reactions in which reactants and products are in different phases. K c = CaCO 3 (s) [CaO][CO ] [CaCO 3 ] CaO (s) + CO (g) [CaCO 3 ] = constant [CaO] = constant K c = [CO ] = K c x [CaCO 3 ] [CaO] K p = P CO The concentration of solids and pure liquids are not included in the expression for the equilibrium constant.

12 Consider the following equilibrium at 95 K: NH 4 HS (s) NH 3 (g) + H S (g) The partial pressure of each gas is 0.65 atm. Calculate K p and K c for the reaction? K p = P NH3 P H S = 0.65 x 0.65 = K p = K c (RT) n K c = K p (RT) - n n = 0 = T = 95 K K c = x (0.081 x 95) - = 1.0 x 10-4

13 General Relations Chemical Kinetics and Chemical Equilibrium k f A + B AB k r rate f = k f [A][B] rate r = k r [AB ] Equilibrium rate f = rate r k f [A][B] = k r [AB ] k f [AB ] = K k c = r [A][B] The equilibrium constant of reversible reaction is the reciprocal of the original equilibrium constant. N O 4 (g) NO (g) NO (g) N O 4 (g) K = [NO ] [N O 4 ] = 4.63 x 10-3 K = [N O 4 ] = 1 [NO ] K = 16

14 General Relations The equilibrium constant of the sum of two or more reactions is given by the product of the equilibrium constants of the individual reactions. A + B C + D C + D E + F A + B E + F K c K c K c K c = [C][D] [A][B] K c = [E][F] [C][D] K c = [E][F] [A][B] K c = K c x K c

15 Writing Equilibrium Constant Expressions The concentrations of the reacting species in the condensed phase are expressed in M. In the gaseous phase, the concentrations can be expressed in M or in atm. The concentrations of pure solids, pure liquids and solvents do not appear in the equilibrium constant expressions. The equilibrium constant is a dimensionless quantity. In quoting a value for the equilibrium constant, you must specify the balanced equation and the temperature. If a reaction can be expressed as a sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions.

16 13.5 Application of the Equilibrium Constant For nonequilibrium conditions, the expression having the same form as K c or K p is called the reaction quotient, Q c or Q p aa + bb + gg + hh + Q c = [C]c [D] d [G] g [H] h The reaction quotient is useful for predicting the direction in which a net change must occur to establish equilibrium

17 13.5 Application of the Equilibrium Constant The reaction quotient (Q c ) is calculated by substituting the initial concentrations of the reactants and products into the equilibrium constant expression. IF Q c = K c the system is at equilibrium Q c < K c system proceeds from left to right to reach equilibrium Q c > K c system proceeds from right to left to reach equilibrium

18 Question The reaction quotient for a system is 7. x 10. If the equilibrium constant for the system is 36, what will happen as equilibrium is approached? a) There will be a net gain in product. b) There will be a net gain in reactant. c) There will be a net gain in both product and reactant. d) There will be no net gain in either product or reactant. e) The equilibrium constant will decrease until it equals the reaction quotient. ANS: b) SECTION: 13.5 LEVEL: medium

19 Calculating Equilibrium Concentrations and Pressure 1. Calculating Q will help in determining the direction of change (x) to establish the equilibrium. Express the equilibrium concentrations of all species in terms of the initial concentrations and a single unknown x, which represents the change in concentration. 3. Write the equilibrium constant expression in terms of the equilibrium concentrations. Knowing the value of the equilibrium constant, solve for x. 4. Having solved for x, calculate the equilibrium concentrations of all species.

20 At C the equilibrium constant (K c ) for the reaction Br (g) Br (g) is 1.1 x If the initial concentrations are [Br ] = M and [Br] = 0.01 M, calculate the concentrations of these species at equilibrium. [ Br ] Q = = =.3 10 > [ Br ] Q>K the system will shift to the left, using x as the change in concentration of Br - K c Br (g) Br (g) Initial (M) Change (M) Equilibrium (M) x -x x x [Br] K c = K [Br ] c = ( x) x = 1.1 x 10-3 Solve for x

21 ( x) K c = = 1.1 x x 4x x = x 4x x = 0 ax + bx + c =0 x = -b ± b 4ac a Initial (M) Change (M) Equilibrium (M) x = 8.74x10-3 Br (g) Br (g) x -x x x x =.98x10-3 At equilibrium, [Br] = x = -5.48x10-3 M or 6.05x10-3 M At equilibrium, [Br ] = x = M

22 13.7 Le Châtelier s Principle If an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset as the system reaches a new equilibrium position. Changes in Concentration N (g) + 3H (g) Equilibrium shifts left to offset stress NH 3 (g) Add NH 3

23 13.7 Le Châtelier s Principle Changes in Concentration continued Remove Add Remove Add aa + bb cc + dd Change Increase concentration of product(s) Decrease concentration of product(s) Increase concentration of reactant(s) Decrease concentration of reactant(s) Shifts the Equilibrium left right right left

24 13.7 Le Châtelier s Principle Changes in Volume and Pressure A (g) + B (g) C (g) Change Increase pressure Decrease pressure Increase volume Decrease volume Shifts the Equilibrium Side with fewest moles of gas Side with most moles of gas Side with most moles of gas Side with fewest moles of gas

25 13.7 Le Châtelier s Principle Changes in Temperature Change Increase temperature Decrease temperature Exothermic Rx K decreases K increases Endothermic Rx K increases K decreases colder hotter

26 13.7 Le Châtelier s Principle Adding a Catalyst does not change K does not shift the position of an equilibrium system system will reach equilibrium sooner uncatalyzed catalyzed Catalyst lowers E a for both forward and reverse reactions. Catalyst does not change equilibrium constant or shift equilibrium.

27 13.7 Le Châtelier s Principle Change Shift Equilibrium Change Equilibrium Constant Concentration yes no Pressure yes no Volume yes no Temperature yes yes Catalyst no no

28 Chemistry In Action Life at High Altitudes and Hemoglobin Production Hb (aq) + O (aq) HbO (aq) K c = [HbO ] [Hb][O ]

29 Chemistry In Action: The Haber Process N (g) + 3H (g) NH 3 (g) H 0 = -9.6 kj/mol

30 QUESTION Consider the following equilibrium: 4NH 3 (g) + 50 (g) 4NO(g) + 6H O(g) What would happen to the system if oxygen were added? 1) More ammonia would be produced. ) More oxygen would be produced. 3) The equilibrium would shift to the right. 4) The equilibrium would shift to the left. 5) Nothing would happen.

31 QUESTION Consider the following system at equilibrium: N (g) + 3H (g) NH 3 (g) kj Which of the following changes will shift the equilibrium to the right? a. Increasing the temperature b. Decreasing the temperature c. Increasing the volume d. Decreasing the volume e. Removing some NH 3 f. Adding some NH 3 g. Removing some N h. Adding some N

32 Review Questions K The equilibrium constant of the following reaction is.18 x 106 H + Br HBr What are the equilibrium concentrations of all the species if the starting concentration of HBr was 0.67 M. (0.67 x) = = x c H + Br HBr Initial (M): Change (M): +x +x x Equilibrium (M): x x (0.67 x) x x = x = The equilibrium concentrations are: [H ] = [Br ] = M [HBr] = 0.67 ( ) = 0.67 M

33 Problem # 4 (or 44 Ed.6) Starting with 4.0 mole of NH3 in.0 L container. After dissociation as given below,.0 mole remains. What is the K for this reaction NH 3 (g) N (g) + 3 H (g) K = 3 3 [N] [H] (0.50)(1.5) K = = [NH3] (1.0) = 1.7 [NH Initial 4.0 mol/.0 L 0 0 Let x mol/l of NH 3 react to reach equilibrium Change -x +x +3x Equil..0 - x x 3x From the problem: [NH 3 ] e =.0 mol/.0 L = 1.0 M =.0 x, x = 0.50 M [N ] = x = 0.50 M; [H ] = 3x = 3(0.50 M) = 1.5 M [N ] [H 3 ] ] 3

34 Problem # 51 At 35 o C, K = 1.6 x 10-5 for the following reaction. Calculate the concentration of all the species, if 1.0 mol of NOCl and 1.0 mole of NO are mixed in 1.0 L flask. 1.6 NOCl(g) NO(g)+ Cl (g) Initial 1.0 M 1.0 M 0 x mol/l of NOCl reacts to reach equilibrium Change -x +x +x Equil x x x 5 10 = ( 1.0+ x) ( x) (1.0) ( x) = (1.0 x) (1.0) (assuming x << 1.0) x = ; Assumptions are great (x is % of 1.0). [Cl ] = M and [NOCl] = [NO] = 1.0 M